Visual interactive differential geometry. Lisbeth Fajstrup and Martin Raussen

Transcription

1 Visual interactive differential geometry Lisbeth Fajstrup and Martin Raussen August 27, 2008

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3 Contents 1 Curves in Plane and Space Vector functions and parametrized curves Parametrization by vector functions Parametrized curves Fundamental properties of smooth vector functions An important consequence Tangents, velocity, and length Secants and tangents Velocity and arc length Exercises Curvature Approximating and osculating circles Principal normals and curvature functions Tangent and normal indicatrix and curvature Acceleration and curvature calculation The Curvature Function characterizes a Plane Curve The local canonical form of a plane curve Exercises Space Curves Moving frame, osculating plane and associated indicatrices Torsion and Frenet s equations Curvature and torsion functions characterize a space curve The local canonical form of a space curve Exercises

4 4 CONTENTS 1.5 Global properties The four-vertex theorem for plane curves Total curvature and Fenchel s theorem Exercises Surfaces in space Surfaces in space - the first definitions About coordinate systems Differentiable functions on surfaces Implicitly given surfaces Regular values of a function Is the solution set of an equation a surface? Particular classes of surfaces Surfaces of revolution Ruled surfaces A useful trick: extending a chart to a local diffeomorphism Curves on a surface Tangent planes and differentials Tangent planes and normal vectors - definitions Tangent planes The differential of a smooth function Normal vectors Oriented Surfaces A Unit Normal Vector Field The Gauss map of an oriented manifold Exercises

5 CONTENTS 5 4 Metric on a surface: The first fundamental form Distances, angles and areas on a surface The metric on the tangent plane The first fundamental form and linear algebra The first fundamental form and arc length The first fundamental form and area Appendix on bilinear forms Bilinear forms. Quadratic forms Bilinear forms and Gram matrices Change of coordinates Diagonalization Curvature functions on regular surfaces Vector fields Normal sections and normal curvature Normal curvature of arbitrary surface curves The Gauss map and its differential Tangent and normal vector fields along a curve Normal curvature and the differential of the Gauss map The second fundamental form Calculating the Weingarten matrix - linear algebra Gauss curvature and mean curvature Classification of points on a surface An outlook at minimal surfaces Isometries. Invariance of the Gauss curvature Isometries between surfaces Theorema egregium and consequences A proof of Theorema Egregium Christoffel symbols Definition The metric determines the Christoffel symbols A second proof of Theorema egregium

6 6 CONTENTS 7 Geodesics and the Gauss-Bonnet theorem Geodesic curvature. Geodesics Definitions The geodesic differential equations Existence and uniqueness The local Gauss-Bonnet theorem Relating geodesic curvature and Gaussian curvature Green s Theorem Hopf s Umlaufsatz Proof: Putting it altogether A note on geodesic triangles The global Gauss-Bonnet theorem The Euler characteristic of a surface Proof of the global Gauss-Bonnet theorem

7 1 Curves in Plane and Space 1.1 Vector functions and parametrized curves You have certainly encountered curves before. Simple examples are straight lines, circles, ellipses; more complex ones the trajectory of a spacecraft or of a charged particle in an electromagnetic field. We want to describe a framework that allows us to describe and investigate general curves in plane and space Parametrization by vector functions Many plane curves can be described as the graph of a function f : [a, b] R. But such a simple curve as a plane circle cannot! And for space curves, it is obvious that one has to find other means of description. Illustration 1.1 graph of a function, circle, and space curve - do it yourself... We let R n = {[x 1, x 2,...,x n ] x i R} denote ordinary n-dimensional { space 1 i = j equipped with the orthonormal basis {e i 1 i n}, e ij = δ ij = 0 i j. Definition 1.2 Let I R denote an open interval. 7

8 8 CHAPTER 1. CURVES IN PLANE AND SPACE 1. Any function r : I R n is called a vector function. A vector function can be described in coordinates as r(t) = i r i (t)e i = [r 1 (t),...,r n (t)], t I. 2. A vector function r : I R n, r(t) = [r 1 (t),...,r n (t)] is called smooth (C ) if the coordinate functions r i (t) are infinitely many times differentiable on I. Its derivative r : I R n is defined as r (t) = [r 1(t),...,r n(t)], t I. Remark 1.3 An interval J can be of the form ]a, b[, ]a, b], [a, b[ or [a, b] for real numbers a, b. One can extend Def. 1.2 and define curves on intervals, which are not open - by requiring r to be continuous on J and smooth on the interior. We may also consider intervals of the form ], b[, ], b], ]a, [, ]a, ] and ], [= R. The interior int(j) removes boundary points from J; f.ex., int([a, b]) =]a, b[. This is needed to extend Def. 1.2 to general intervals, since differentiability only makes sense on open intervals! Parametrized curves Let us get started with some examples: A line in Euclidean n-space through P in direction x R n \ {0} can be parameterized by the vector function r : R R n, r(t) = OP t = OP + tx, t R. The vector function c : [0, 2π] R 2, c(t) = [cost, sin t], represents a circle C in the Euclidean plane with radius 1 and the origin as its center: The circle consists of all points P t with OP t = c(t). In both cases, you may imagine the arrow OP t pointing at P t at time t. Remark, that a circle cannot be represented as the graph of a function f : R R: There are two elements y = ± 1 x 2 corresponding to an element x ( 1, 1) with [x, y] C. In space, it is even less reasonable to represent curves as graphs of functions. Example 1.4 The vector function r : R R 3, r(t) = [a cost, a sin t, bt] represents a helix winding around a cylinder of radius a with the z-axis as the central axis above, resp. below a circle of radius a. The helix will be used as one of our central examples throughout this chapter.

9 1.1. VECTOR FUNCTIONS AND PARAMETRIZED CURVES Figure 1.1: A helix Illustration 1.5 Better have an applet here: An interval, the circle/helix, and a scrollbar allowing to scroll a point one the interval, the corresponding point on the helix, and an arrow from the origin to that point (moving when scrolled!) See the geometric lab. Definition 1.6 A smooth vector function r : I R n is called a parametrization of the curve C = {P t R n OP t = r(t), t I} The curve C consists thus of all the points P t pointed to by arrows OP t = r(t). Often, you will imagine I as a time interval, and the curve given by a particle at position P t at time t. Example 1.7 (important general example): Is it always possible to represent a the graph of a function f : I R in the plane by a (vector function) parametrization? Yes! Here is the recipe: The curve C = {[x, f(x)] R 2 x I} can be parameterized by the vector function r : I R 2, r(t) = [t, f(t)], t I. Then, the points P t : [t, f(t)] run through all the points on the curve C. For instance, r(t) = [t, sin t] is a parametrization of the graph of the sine-function. Illustration 1.8 Scrollable interval and arrows to the graph.

10 10 CHAPTER 1. CURVES IN PLANE AND SPACE Fundamental properties of smooth vector functions Since we shall express geometric properties of curves using vector functions and their derivatives (up to degree 3), a brief investigation into properties of such vector functions will prove to be a good investment. The following rules apply to vector functions and ordinary functions that can be described as composites (sums, products, dot products, cross=wedge-products, composite of maps) of vector functions (and ordinary functions). Make sure that you understand the meaning of each term! Proposition 1.9 Let I, J R denote intervals; moreover, let r 1,r 2 : I R n denote smooth vector functions, and let f : I R and s : J I denote (ordinary) smooth functions. The derivatives of compound functions satisfy the following rules at every t int(i): 1. (r 1 ± r 2 ) (t) = r 1 (t) ± r 2 (t); 2. (fr 1 ) (t) = f (t)r 1 (t) + f(t)r 1 (t); 3. (r 1 r 2 ) (t) = r 1 (t) r 2(t) + r 1 (t) r 2 (t); 4. (r 1 r 2 ) (t) = r 1 (t) r 2(t) + r 1 (t) r 2 (t); 5. (The chain rule) (r 1 s) (t) = s (t)r 1(s(t)), t J. Proof: Having described vector functions by their coordinates, the proofs are straightforward implications of the rules for derivatives of ordinary smooth functions. Note, that the rules for the dot product (3.) and the cross product (4.) have the form of the ordinary product rule, again. Note that the terms in (3.) are ordinary smooth functions! An important consequence The following consequence of the derivation rules above is completely elementary, but nevertheless a technically very important device: Proposition 1.10 (The fundamental trick). Let c R denote a constant, and I R an interval. 1. Let r 1,r 2 : I R n denote two smooth vector functions with

11 1.2. TANGENTS, VELOCITY, AND LENGTH 11 r 1 (t) r 2 (t) = c for every t I. Then, r 1(t) r 2 (t) + r 1 (t) r 2(t) = 0 for every t int(i). 2. Let r : I R n denote a smooth vector function with constant length r(t) = c for every t I. Then r(t) r (t) = 0 for every t int(i). In particular, r (t) is perpendicular on r(t) for every t int(i). Proof: Apply the product rule (1.9.3) for vector functions. 1.2 Tangents, velocity, and length Secants and tangents For the graph of a function f : I R, the secant through [x 0, f(x 0 )] and [x 1, f(x 1 )] with x 0 x 1 is defined as the line through these two points on the graph. The tangent to the graph at [x 0, f(x 0 )] is defined as the limit position of these secant lines when x 1 tends to x 0 if such a limit position exists. We know that it exists when f is smooth; in that case it is the line through [x 0, f(x 0 )] with slope f (x 0 ). It is not very difficult to generalise these concepts to curves (in the plane, in 3-space or in n-space) given by a smooth parametrization. In fact, smoothness is not quite enough to guarantee the existence of a tangent line at every point. This can be seen in Example 1.11 Let C denote the curve given by the parametrization r : R R 2, r(t) = [1 + t 2, 1 + t 3 ], cf. Fig Apparently, this curve has a singularity at the point OP 0 = [1, 1]. The secant lines do have a limit position, which is the horizontal line through P 0. But this line is not a good approximation to the curve near P 0. Points in the 2. quadrant in the line do not approximate the curve. Illustration 1.12 Scroll on the tangent line and on the curve using the differential, both at P 0 and at P 1. But in most cases, a tangent line yields a good approximation, e.g., at the point P 1 : [2, 2] for the curve above. Why do we find different behaviour? Let us calculate derivatives: r (t) = [2t, 3t 2 ]; r (0) = 0 r (1) = [1, 1] 0!

12 12 CHAPTER 1. CURVES IN PLANE AND SPACE Figure 1.2: A curve with a singular point Definition A smooth vector function r : I R n is called a regular parametrization of the curve C = {P t R n OP t = r(t), t I} if and only if r (t) 0 for all t int(i). 2. A subset C R n is called a regular curve if there is a regular parametrization r : I R n with C = {P t R n OP t = r(t), t I}. Remark There is nothing like the parametrization of a curve. You can move along a curve at different speed, and speed need not be constant. 2. A regular curve may have a non-regular parametrization. This is the case for the circle with the parametrization c(t) = [cost 2, sin t 2 ], t R. Why? 3. The curve C from Ex does not possess any regular parametrization. Tangents of regular curves Given a (regular) curve C with a regular parametrization r : I R n, it is easy to show that it has a tangent line at every point P t C, t int(i) and to determine it: Given a point P t0 C with r(t 0 ) = OP t0, t 0 int(i). To get the tangent line to C at P t0, we only need a parallel vector as a limit vector from suitable parallel vectors of secant lines nearby. The secant between P t0 and P t with r(t) = OP t has parallel vector P t0 P t = OP t OP t0 = r(t) r(t 0 ) (for P t0 P t ). For t t 0,

13 1.2. TANGENTS, VELOCITY, AND LENGTH 13 this difference vector gets shorter and shorter and thus tends to 0, hence does not give information on the direction of a possible tangent. Instead, we look at the unit parallel vector P t0 P t = r(t) r(t 0) P t0 P t r(t) r(t 0 ) and its limit position for t t 0. Illustration 1.15 Applet with secants and tangents See the geometric lab.. Proposition 1.16 Let r : [a, b] R n denote a regular parametrization of a curve C. Let t 0 (a, b) and OP t0 = r(t 0 ). Then, the curve has a tangent line at P t0 with parallel vector r (t 0 ). A parametrization for the tangent line is t(t) = r(t 0 ) + tr (t 0 ). Proof: For a regular parametrization, we take the limit of the unit direction vectors for t t 0 + from the left: P t0 P t lim t t 0 + P t0 P t = lim r(t) r(t 0 ) t t 0 + r(t) r(t 0 ) = lim t t 0+ and similarly, r(t) r(t 0 ) t t 0 lim t t0 + r(t) r(t 0) P t0 P t lim t t 0 P t0 P t = r (t 0 ) r (t 0 ). = r (t 0 ) r (t 0 ), t t 0 (Why does this not work for r (t 0 ) = 0?). In particular, these two limit vectors are the two unit vectors parallel to r (t 0 ). Hence the line through P t0 with parallel vector r (t 0 ) is a limit for the secant lines through P t0 and thus rightly qualifies as the tangent line. Illustration 1.17 Applet firing tangents from points at a regular curve given by a parametrization after choice. Including coordinates of position and direction. Remark that a regular curve has a tangent vector at each point. On the other hand, a curve with a non-regular parametrization may have tangent lines, but Ex shows that there may be points without a tangent line (regardless the parametrization). In many cases, one can still define semi-tangents using the limits as the definition. lim t t 0 + r(t) r(t 0 ) r(t) r(t 0 ), resp. lim t t 0 r(t) r(t 0 ) r(t) r(t 0 )

14 14 CHAPTER 1. CURVES IN PLANE AND SPACE Velocity and arc length Speed and unit tangent vectors Proposition 1.16 shows that the derivative r (t) of a regular parametrization of a curve C contains the geometric information needed to determine the tangent lines to C. But there is more (non-geometric) information hidden in the vector function r (t). If you imagine r(t) tracing the trajectory of a particle at time t, differentation should have something to do with the speed of that motion, as well. How can one grasp the speed of a smooth vector function? As for functions of one variable, we look at the quotient P t0 P t. For t t 0, we obtain t t 0 = r(t) r(t 0) t t 0 the limit lim t t0 r(t) r(t 0 ) t t 0 = lim t t0 r(t) r(t 0 ) t t 0 = r (t 0 ). Definition 1.18 Let r : I R n denote a smooth vector function. At t 0 int(i), it has speed v(t 0 ) = r (t 0 ) and unit tangent vector t(t 0 ) = r (t 0 ) r (t 0. In particular, ) r (t 0 ) = v(t 0 )t(t 0 ) for all t 0 int(i). Illustration 1.19 Parametrization: Interval, derivatives, tangents and speed curve. Reparametrizations How are different parametrizations of the same regular curve related to each other? Well, they should give rise to the same tangent lines along the curve, whereas speed functions can be entirely different. Here is a method to reparametrize a regular curve with a given regular parametrization r : I R n : Choose a (strictly monotone) smooth bijective 1 function f : J I from an interval J such that f (t) > 0 for all t int(j). Then the composite vector function r 1 = r f : J R n parametrizes C since r 1 (t) = r(f(t)) or r(u) = r 1 (f 1 (u)), and it is regular (Why? The chain rule in Prop ) Illustration 1.20 Two intervals, a reparametrization, and the motion along the curve. (three moving spots) We use this approach to make the definition of a regular curve from Def more precise. Remark first that we may define an equivalence relation on regular parametrisations: r r 1 if and only if r 1 is a reparametrization of r; cf. Exc and onto

15 1.2. TANGENTS, VELOCITY, AND LENGTH 15 Definition 1.21 A regular curve is an equivalence class of regular parametrisations under the equivalence class above. Since we only allow reparametrization functions f with f > 0 everywhere, we can speak about an orientation (order) on the curve: r(t 0 ) comes before r(t 1 ) if and only if t 0 < t 1. Arc Length Let r : I R n denote a smooth vector function that is a regular parameterisation of a regular curve C R n. Definition 1.22 The length l(c) of the curve C is defined as the non-negative real number I r (t) dt. For r(t) = [x 1 (t),...,x n (t)], we obtain: l(c) = (x 1 (t)) (x n (t))2 dt. I Example For the helix from Ex. 1.4 with parametrization r : R R 3, r(t) = [a cost, a sin t, bt], we calculate: r (t) = (a sin t) 2 + (a cost) 2 + b 2 = a 2 + b 2. Hence, one turn around the helix has length l = s(2π) = 2π 0 a2 + b 2 dt = 2π a 2 + b 2. Remark that you obtain the familiar formula for the arc length of a circle in case b = For the graph of a function f : [a, b] R from Ex. 1.7 with parametrization r : [a, b] R 2, r(t) = [t, f(t)], a t b, we calculate the length l of the graph as follows: r (t) = [1, f (t)] = 1 + (f ) 2 (t), and hence l = b a 1 + (f ) 2 (t)dt. The last example shows, that it is very often difficult or impossible to calculate the length of a curve in explicit terms. Many of the integrands involving square roots do not have explicit antiderivatives!

16 16 CHAPTER 1. CURVES IN PLANE AND SPACE Here is a motivation for Def. 1.22: For t Ī, let I t = I ], t] denote the part of the interval containing numbers less than or equal to t. Then, the vector function r t : I t R n, r t (u) = r(u) is a regular parametrization of a section C t of the curve C. Let s(t) denote the length of C t. The function s : I R is called the arc length function of the parametrization. Note that s is monotonous: Increasing t yields increasing s(t)! Arc length s(t) and speed v(t) of the parametrization r should be related by the property: s (t) = v(t) = r (t), and hence s(t) = I r (u) du + c, c a real constant. For t 0 = inf I, the length of the empty curve C t0 has to be zero, hence c = 0. Illustration 1.24 Scrollable interval, parametrization, section C t, functions s(t) and s (t). If the concept of length has any value, it should be independent of the choice of a regular parametrization of a given curve. This is indeed the case; see Exc Unit Speed Parametrization It is somehow awkward to have too many parametrizations for the same curve. For theoretical purposes, one needs a particular nice one: Definition 1.25 A vector function r al : I R n is called a unit speed parametrization if its speed satisfies v(s) = r al (s) = 1 for every s int(i). In other words, the derivative of r al is the unit tangent vector t(s) at every element s I : t(s) = r al (s). It has the following property which explains its second name an arc length parametrization: Proposition 1.26 For t 0 t 1 I, the arc length function s associated with a unit speed parametrization has the property: s(t 1 ) s(t 0 ) = t 1 t 0. Proof: s(t 1 ) s(t 0 ) = I t1 r al (s) ds I t0 r al (s) ds = t 1 t 0 r al (s) ds = t 1 t 0 ds = t 1 t 0.

17 1.2. TANGENTS, VELOCITY, AND LENGTH 17 Example 1.27 The parametrization r(t) = [a cost, a sin t, bt] for the helix from Ex. 1.4 yields (cf. Ex. 1.23): v(t) = r (t) = a 2 + b 2 ; in particular, it is a constant function. Hence, starting at t = 0, the arc length function is given as s(t) = a 2 + b 2 t; and thus its inverse is t(s) = s a. Substituting t(s) into the 2 +b parametrization r yields the arc length parametrization 2 for the helix, to wit: r al (s) = r(t(s)) = [a cos s s a sin a2 + b2, a2 + b 2, b s a2 + b 2]. Remark The following gives an easy intuitive idea for the arc length parametrization of a given curve: Imagine a piece of rope with a scale (starting at 0) and bend it along the curve (without stretching!) Then, r al (s) is the vector from the origin to the point corresponding to the mark s on the scale. 2. The advantage of the arc length parametrization is that it focusses on the geometric properties of the curve rather than the infinitely many possible different modes (with varying speed, acceleration etc.) to run through it. Using this parametrization, it will be much easier to define entities like the curvature; on the other hand, for concrete calculations, one usually does not dispose of a concrete arc length parametrization. Does a regular curve always admit a unit speed parametrization? Well, in most cases, it is not possible to write down an explicit unit speed parametrization using familiar functions as components. Nevertheless, a unit speed parametrization always exists: To find it, let us first assume that there is a unit speed parametrization r al : I R n representing a curve C. Let f : J I denote any bijection with f (t) > 0 giving rise to the (re)parametrization r = r al f of the curve C. The two parametrizations are linked via the arc length function s : J [0, l(c)] associated to the parametrization r, to wit: r(t) = r al (s(t)). ( At time t, we trace the point at distance s(t) from the start point ). How could one (re)construct r al given r? Well, if the function s had an inverse t : [0, l(c)] J, then r al (s) = r al (s(t(s))) = r(t(s)). (1.1) Illustration 1.29 Scrollable s and t-intervals and parametrization.

18 18 CHAPTER 1. CURVES IN PLANE AND SPACE The function s : J R, s(t) = I r (t) dt has the derivative s (t) = r (t) = f (t) > 0, and hence s is strictly increasing. Therefore, it has in fact a smooth inverse, and we can use (1.1) as the definition of an arc length (unit speed) parametrization. Remark that we have shown the existence of a unit speed parametrization using the existence of a function t inverse to s. In most cases, it is not possible to give a formula for t(s) in terms of well-known functions. Just try for the graph of a differentiable function (cf. Ex. 1.7). As you may have observed by now, it is customary to use the parameter s for an arc length parameter, and t for an arbitrary parameter. Whenever s is used in this text, we talk about a unit speed parametrization. A geometric property of unit speed parametrizations The following property of unit speed parametrizations is the main reason why geometers like them so much. It is an immediate consequence of the fundamental trick (Prop. 1.10): Proposition 1.30 Let r : I R n denote a unit speed parametrization. Then, r (s) r (s) = 0 for all s int(i). In geometric terms, the acceleration vector r (s) is always perpendicular on the tangent vector r (s) Exercises 1. Calculate tangents of several curves and use applet to show them. 2. Determine a parametrization for the tangent line to the graph of the function f :]a, b[ R (with parametrization r :]a, b[ R 2, r(t) = [t, f(t)] at t 0 ]a, b[. Is the result familiar? 3. How can you determine from r (t 0 ) whether the curve C has a horizontal/vertical tangent at P t0? 4. Why can the curve from Ex not have any regular parametrization at P 0 : [1, 1]? 5. Show that the relation on regular parametrisations (r r 1 if and only if there is a bijective function f : J I with f (t) > 0 for every t int(j)

19 1.3. CURVATURE 19 with r 1 = r f) is an equivalence relation (i.e., reflexive, symmetric, and transitive). One may change the definition of by allowing bijective reparametrisations f with f (t) 0 for every t int(j). How is the geometric meaning changed? 6. Show that the length of a regular curve C is independent of the choice of a regular parametrization. (Hint: Let r : I R n denote a regular parametrization of C, let f : J I denote a real function such that r 1 = r f : J R n is a regular reparametrization of C. Using the chain rule for derivatives and substitution for integals, show, that r 1 du = r dt.) J 7. Find a unit speed parametrization of the regular space curve C given by the parametrization r(t) = [t, 6 2 t2, t 3 ], t R. I 1.3 Curvature From now on, we restrict ourselves to (regular) plane and space curves, given by regular smooth parametrizations r : I R n, n = 2 or n = 3. What should the curvature of a curve be? Well, a line is not curved at all; its curvature has to be zero. A circle with a small radius is more curved than a circle with a large radius. Circles and lines have constant curvature. Curves that are not (pieces of) circles or lines will have a curvature varying from point to point Approximating and osculating circles A tangent line to a curve C can be defined as the limit of the secant lines through two points on the curve. In much the same spirit, one can define curvature circles (= osculating circles) as the limit circles of circles passing through three points on the curve. The curvature of C at a point P is then defined as the inverse of the radius of this curvature circle at P. Let C denote a plane or space curve given by a unit speed parametrization r : I R n, n = 2 or n = 3. Let s 0 int(i) and P s0 the point on C with r(s 0 ) = OP s0.

20 20 CHAPTER 1. CURVES IN PLANE AND SPACE To construct the curvature circle at P s0, choose s 1 and s 2 close to s 0 such that P s0, P s1 and P s2 are not contained in a line (This is possible unless the curve is coinciding with a line close to P s ). In Exc , you are invited to show that there exists exactly one circle through the points P s0, P s1 and P s2 ; we denote its center by C(s 0, s 1, s 2 ). The idea is to find a limit circle when both s 1 and s 2 approach s 0. To this end, we investigate the real smooth function d(s) = (r(s) C(s 0, s 1, s 2 )) (r(s) C(s 0, s 1, s 2 )). Its geometric meaning is the square of the distance between P s and C(s 0, s 1, s 2 ). By definition, d(s 0 ) = d(s 1 ) = d(s 2 ), i.e., the square of the radius of the circle through these three points. By Rolle s theorem for smooth functions in one variable, there exist two intermediate values t 1, t 2 such that d (t 1 ) = d (t 2 ), and, moreover an intermediate value u such that d (u) = 0. The derivatives of d can be calculated as d (s) = 2r (s) (r(s) C(s 0, s 1, s 2 )), resp. (1.2) d (s) = 2(r (s) (r(s) C(s 0, s 1, s 2 )) + 2r (s) r (s) = 2(r (s) (r(s) C(s 0, s 1, s 2 )) + 2. (1.3) To obtain the last equality, we used that r is a unit speed parametrization. At s = u, we obtain in particular: r (u) (r(u) C(s 0, s 1, s 2 )) = 1. Illustration 1.31 Circles through 3 points on the curve converging to osculating circle. The function d and intermediate points. See geometric lab. Now, let s 1 and s 2 tend to s 0 to get a limiting curvature circle with centre C(s 0 ) and radius ρ(s 0 ). You are invited to show the existence of this curvature circle in Exc When s 1 and s 2 tend to s, t 1, t 2 and u tend to s 0, as well. This allows us to derive from (1.2), that r (s 0 ) (r(s 0 ) C(s 0 )) = 0, and that r (s 0 ) (r(s 0 ) C(s 0 )) = 1. We obtain the following geometric implications: Lemma 1.32 The tangent line to C at P s0 (parallel to t(s 0 ) = r (s 0 )) is perpendicular to the radial vector P s0 C s0 and, for a plane curve The acceleration vector t (s 0 ) = r (s 0 ) is perpendicular to r (s 0 ) according to Prop. 1.30, and hence, it has to be parallel to r(s 0 ) C(s 0 ); calculating lengths, we obtain: r (s 0 ) r(s 0 ) C(s 0 ) = 1, i.e., t (s 0 ) = r (s 0 ) = 1 ρ(s 0. The length of the acceleration vector is thus inverse to the radius of ) curvature at s 0.

21 1.3. CURVATURE 21 This length t (s 0 ) = r (s 0 ) is the absolute value of the curvature of the plane curve C at P s0. We shall introduce a refinement, the signed curvature for plane curves in the next Sect Illustration 1.33 Curve, r (s 0 ),r(s 0 ) C(s 0 ),r (s 0 ), r (s 0 ). Scrollable. For a space curve C, one can follow the same line of argument; we need the existence of the osculating plane and its properties to complete the description of the curvature for space curves. See Sect Principal normals and curvature functions Let C be a regular smooth curve in the plane or in 3-space with unit speed parametrization r : I R n, n = 2 or 3. By Def and its explanation, the vector t(s) = r (s) is a unit tangent vector to the curve at the point P s. The vector function t : I R i is called the unit tangent vector field moving along the curve. We are now going to analyse the information hidden in the derived vector field t = r along the curve C. An application of Prop yields: Corollary At every point P s of the curve, the derivative t (s) is perpendicular to t(s) : t (s) t(s) = For a plane curve C the vectors t (s) and ˆt(s) are parallel. Definition The vector t (s) is called the curvature vector at the point P s on C with OP s = r(s). 2. A point P s on C is called an inflection point if t (s) = The principal normal vector n(s) to the curve at P s is defined as follows: (a) For a plane curve C let n(s) = ˆt(s). (b) For a space curve C let n(s) = t (s) t (s) at every non-inflection point P s C. Definition 1.36 The curvature κ(p s ) of the curve C at the point P s is defined as follows: 1. For a plane curve C let and thus κ(s) = ± t (s). t (s) = κ(s)n(s), (1.4)

22 22 CHAPTER 1. CURVES IN PLANE AND SPACE 2. For a space curve C let κ(s) = t (s) 0. (1.5) and thus for κ(s) 0 : t (s) = κ(s)n(s). C n t P ρn ρn n Q t Figure 1.3: Tangent and principal normal vectors and osculating circles at points P and Q. The radius of curvature is ρ = 1/κ Remark By Cor. 1.34, tangent and principal normal vectors are perpendicular to each other at every point of the curve: t(s) n(s) = An inflection point P s on C is characterised by the property κ(s) = 0. Illustration 1.38 Moving tangent, t (s),n(s), osculating circle; view inflection points; moreover, plot curvature. Here are several motivations for the definition of the curvature above: 1. The magnitude t (s) of the derivative t (s) measures the rate of change of the direction of the tangent vector field t(s) since its length t (s) is constant. The faster the direction of the tangent vector changes, the more curved is the curve.

23 1.3. CURVATURE Let us calculate the curvature for a plane circle with radius R. With center at the origin, the arc length parametrization of the circle is given by r al (s) = [R cos s, R sin s ]. We calculate: R R t(s) = [ sin s R, cos s R ], t (s) = [ 1 R cos s R, 1 R sin s R ] = 1 Rˆt(s), and according to (1.4), κ(s) = 1 for every s. Hence, the curvature of a R circle is constant and in inverse proportion to its radius as it should! For a general curve C, the curvature at a point P is in inverse proportion to the radius of the best approximating circle at P, the so-called osculating circle, cf. Fig. 1.3 and Sect Tangent and normal indicatrix and curvature In the plane, a unit vector field v : I R 2 takes values in the unit circle S 1 := {[x, y] R 2 x 2 + y 2 = 1} = {[cos(θ), sin(θ)] θ R}. It can thus be seen as a continuous map v : I S 1 ; if the curve does not have double points 2, one may view it as a map from C to S 1, the so-called indicatrix. We shall later use a similar point of view to define the Gauss map on a surface. Both the unit tangent vector field t(s) and the principal normal vector field ˆt(s) give rise to (perpendicular) tangent and normal indicatrices. Illustration 1.39 Tangent and normal indicatrices of a plane curve; simultaneously moving points on interval, curve and S 1. To relate to curvature calculations, we choose a description of the unit tangent vector field as t(s) = [cosθ(s), sin θ(s)] with θ(s) the angle between t(s) and the horizontal vector i, i.e., cosθ(s) = t(s) i 3. A calculation of t (s) using the chain rule (Prop. 1.9(5)) yields: t (s) = θ (s)[ sin θ(s), cosθ(s)] = θ (s)ˆt(s), (1.6) and hence: κ(s) = θ (s). Hence, the curvature measures the rate of change for the angle between tangents, as it should. Moreover, we get an explanantion for the sign of the curvature of a plane curve, to wit: Corollary 1.40 Near the point P s, a plane curve C is curved 2 and neither converges to a point already on the curve at an end of the interval 3 This can easily be done locally and then lifted along the entire curve. Try Exc

24 24 CHAPTER 1. CURVES IN PLANE AND SPACE counter-clockwise if and only if κ(s) > 0, and clockwise if and only if κ(s) < 0. In Fig. 1.3, the curvature is negative at P and positive at Q. Illustration 1.41 Banchoff s curvature caterpillar. One needs to explain, that n (s) = κ(s)t(s) Exc ?! Indicatrices for space curves will be covered in Sect Acceleration and curvature calculation The definition of curvature above uses the unit speed parametrization of a given curve. But in general, you have only a regular parametrization r : J R n at hand. In this case, the curvature at a given point P is hidden in the acceleration vector at that point. Let us first compare the derivatives of different parametrizations of the same curve C: Let r : J R n denote a regular smooth parametrization, let s : J I = s(j) R denote the associated arc length function s(t) = I t r (u) du. Let r al : I R n the unit speed (arc length) reparametrization (cf. Sect. 1) with the property r(t) = r al (s(t)). Let us calculate the derivatives of the parametrization r(t) using the chain rule and the product rule from Prop. 1.9 and the definition of curvature from Def. 1.36: r (t) = s (t)r al (s(t)) = v(t)t(t); r (t) = v (t)r al(s(t)) + (s (t)) 2 r al(s(t)) = v (t)t(t) + v(t) 2 κ(t)n(t). (1.7) (If P t C R n is the point with OP t = r(t), then v(t) is the speed at P t associated to the parametrization, a(t) = v (t) = s (t) is the scalar acceleration at P t, t(t) is the unit tangent vector to C at P t, κ(t) the curvature of C and n(t) the principal normal vector to C at P t.) Before using (1.7) to calculate the curvature of a given curve, let us look at the following attractive interpretation in mechanics: Equation (1.7) yields a decomposition of the acceleration vector a(t) = r (t) into a tangential component a t (t) and a normal component a n (t) : a(t) = a t (t) + a n (t) = v (t)t(t) + v 2 (t)κ(t)n(t).

25 1.3. CURVATURE 25 In particular, the magnitude of the tangential component is: a t (t) = v (t), which is the scalar accelaration, i.e., the rate of change of the speed. The magnitude of the normal component is: a n (t) = v 2 (t)κ(t). Hence, the force acted upon a particle normal to its path is proportional to the square of its speed and to the curvature of the curve. This is intuitively known to every car driver; when you drive through a narrow curve, you have to slow down drastically in order to avoid strong normal forces. Curvature formulas Proposition Let C be a plane curve with parametrization r : J R 2. Its curvature κ(t) at a point P t with OP t = r(t) is given by κ(t) = [r (t),r (t)] r (t) 3. (1.8) The numerator [r (t),r (t)] is the plane product of the vectors r (t),r (t) R 2. More explicitly, for r(t) = [x(t), y(t)], t J, we obtain: x (t) x (t) y (t) y (t) κ(t) = ( (1.9) x (t) 2 + y (t) 2 ) Let C be a space curve with parametrization r : J R 3. Its curvature κ(t) at a point P t with OP t = r(t) is given by Proof: κ(t) = r (t) r (t) r (t) 3. (1.10) More explicitly, for r(t) = [x(t), y(t), z(t)], t J, we obtain: κ(t) = [x (t), y (t), z (t)] [x (t), y (t), z (t)] ( x (t) 2 + y (t) 2 + z (t) 2 ) 3. (1.11) 1. Using properties of the determinant of a (2 2)-matrix and (1.7), we obtain: [r (t),r (t)] = [v(t)t(t), v (t)t(t)+v 2 (t)κ(t)ˆt(t)] = [v(t)t(t), v 2 (t)κ(t)n(t)] = v 3 (t)κ(t)[t(t),ˆt(t)] = v 3 (t)κ(t). (The last equation uses that the plane product of a unit vector and its hat vector is one, since it measures the (signed) area of the rectangle spanned by both vectors.)

26 26 CHAPTER 1. CURVES IN PLANE AND SPACE 2. Using properties of the cross product and (1.7), we obtain: r (t) r (t) = v(t)t(t) (v (t)t(t)+v 2 (t)κ(t)n(t)) = v(t)t(t) v 2 (t)κ(t)n(t) = v 3 (t)κ(t)(t(t) n(t)). The vector t(t) n(t) is a unit vector, since t(t) and n(t) are mutually orthogonal unit vectors. Therefore, the vector r (t) r (t) has length v 3 (t)κ(t). Example An ellipse E with semi-axes a and b has a parametrization r(t) = [a cost, b sin t], t [0, 2π]. We calculate: r (t) = [ a sin t, b cost], r (t) = [ a cos t, b sin t], [r (t),r (t)] = a sin t a cost b cost b sin t = ab. Since r (t) = a 2 (sin t) 2 + b 2 (cost) 2, we obtain: κ(t) = ab. (1.12) (a 2 (sin t) 2 + b 2 (cost) 2 ) The curve C f given as the graph of a function f : [a, b] R (Ex. 1.7) can be parameterised by the vector function r : [a, b] R 2 given as r(t) = [t, f(t)]. We calculate: r (t) = [1, f (t)], r (t) = [0, f (t)]; hence, [r (t),r (t)] = f (t), r (t) = 1 + (f (t)) 2, and κ(p t ) = f (t) (1.13) 1 + (f (t)) 23. In particular, C f is curved counter-clockwise at P t if f (t) > 0 and clockwise if f (t) < 0 (Cor. 1.40). 3. Let r denote the parametrization for a helix from Ex. 1.4 given as r(t) = [a cost, a sin t, bt], a, b > 0. Its derivatives (velocity and acceleration vectors) are calculated as r (t) = [ a sin t, a cost, b]; r (t) = [ a cos t, a sin t, 0]. (1.14)

27 1.3. CURVATURE t Figure 1.4: Ellipse and curvature function Hence, r (t) = a 2 + b 2 ; r (t) r (t) = [ab sin t, ab cost, a 2 ]; r (t) r (t) = a a 2 + b 2. (1.15) The curvature of the helix at P t is calculated as κ(t) = a a 2 + b 2 a2 + b 23 = a a 2 + b2. (1.16) Note, that the curvature is constant along the helix. For b = 0, we get as a special case the curvature 1 of a circle with radius a. a Osculating Circles In Sect , we motivated the curvature concept via the radii of osculating circles approximating the curve. Having determined the curvature of a curve with a given parametrization, we can now determine the osculating circles for a plane curve C given by a regular smooth parametrization r : I R 2.

28 28 CHAPTER 1. CURVES IN PLANE AND SPACE Proposition 1.44 The osculating circle at P t with OP t = r(t) has radius ρ(t) = 1 and center C κ(t) t with P t C t = 1 n(t). κ(t) Proof: Exc Use Lemma In a similar way, one can determine the osculating circles along a space curve. In that case, one has to use that the osculating circle through P t is contained in the osculating plane ω(t) through P t (cf. Sect ). The Evolute Curve From a given curve C, one may obtain a new curve E C, the evolute of C, by associating to every point P C the corresponding centre of curvature C P, i.e., the center of the osculating circle through P. It is easy to translate the description of this center in 1.44 into a parametrization of E C : Corollary 1.45 Let C be a curve with parametrization r : I R 2. The following is a parametrization for the evolute E C of C: e(t) = r(t) + n(t) κ(t) = r(t) + r (t) 3 [r (t),r (t)] r (t) r (t) = r(t) + r (t) 2 [r (t),r (t)] r (t). Example The evolute of a cycloid is a translated cycloid. 2. The evolute of an ellipse is an astroid. Using Cor. 1.45, we obtain the parametrization e(t) = [ a2 b 2 (cost) 3, b2 a 2 (sin t) 3 ]. a b Illustration 1.47 Replace figure by an applet producing moving osculating circles and the evolute (as done by Robert for a specific curve) Proposition 1.48 Let C denote a regular plane curve with constant curvature κ. For κ = 0, C is a straight line segment; for κ 0, C is contained in a circle. Proof: Let r denote a unit speed parametrization of C. If κ = 0, then r = 0 (why?), and r (s) is a constant vector v. Hence, r(s) = sv +b for some constant vector b; the image of an interval is thus contained in a straight line.

29 1.3. CURVATURE Figure 1.5: Ellipse with osculating circles and evolute Now suppose κ 0. The evolute curve of a circle is a constant point e(t) = e. Let us first show, that the evolute curve of C is constant, too: By Cor. 1.45, it is given by e(s) = r(s) + n(s) κ. Hence, e (s) = r (s) + n (s) κ = t(s) κ t(s) = 0. κ (The middle equation stems from Exc (c)). As a result, e(s) is a constant vector e, and r(s) e = n(s) has constant length r(s) e = 1. Hence, the κ κ image of r is contained in the circle with center at e and radius 1. κ The Curvature Function characterizes a Plane Curve We have seen how to associate to a curve its curvature function. Illustration 1.49 Again. Curve and asociated curvature function. Is it possible to somehow reverse this process? More precisely: Let k : I R denote a continuous function. Is there always a plane curve C (with unit speed

30 30 CHAPTER 1. CURVES IN PLANE AND SPACE parametrization r : I R 2 ) realizing this function as the curvature function associated to C? How many such curves are there? The answer is, that there is always such a curve, and it is uniquely determined up to a rigid motion of the plane, i.e., up to a combination of a translation and a rotation. First of all, it is clear, that a translation, resp. a rotation does not change the curvature function κ. On the other hand the following iterative construction shows why one should expect the existence of a curve with given curvature function 4 : Choose a start point P 0 and a start direction given by a unit vector v 0 (this choice corresponds to the choice of a translation and a rotation). Draw a circular arc of length s through P 0 with tangent vector v 0 and radius ρ 0 = 1. At the end of κ(0) this little circle you obtain a point P s and a tangent direction v s. Continue with a circular arc of length s through P s with tangent vector v s and radius ρ s = 1 κ(s) and obtain an end point P 2s and an end direction v 2s. Keep on. The resulting curve will not be smooth everywhere, but for small s, the curvature will be a step function close to the original κ. Before the computer age, this method was in fact sometimes used to graph a curve given by a parametrization! Illustration 1.50 Curve consisting of circular arcs and curvature step function Let us now transform this intuitive reasoning into a theorem with a strict proof: Let I R denote an open interval, let C k (I,R n ) denote the continuous (vector) functions r : I R n that are k times differentiable on I with continuous derivatives 5. Associating to a unit speed parametrization of a plane curve its curvature function corresponds to a map K : {r C 2 (I,R 2 ) r (s) = 1 for all s I} C 0 (I,R). A translation of the answer given initially into technical terms reads as follows: Theorem 1.51 (The fundamental theorem for plane curves) 1. The map K is onto. 2. K(r 1 ) = K(r 2 ) if and only if there exists a rigid motion Φ : R 2 R 2 such that r 2 = Φ r 1. 4 Another way to introduce this construction is: How to find a curve whose curvature function is a step function 5 Should we introduce this notation earlier?

31 1.3. CURVATURE 31 Proof: The map K can be decomposed as K : {r C 2 (I,R 2 ) r C 1 (I, S 1 )} D 1 C 1 (I, S 1 ) D 2 C 0 (I,R) with D 1 denoting differentiation. To describe the map D 2, note first that every vector function t C 1 (I,S 1 ) can be parameterized by a C 1 - (angle) function θ : I R such that t(s) = [cosθ(s), sin θ(s)] with θ well-determined up to a constant 2πn, n Z (cf. Sect ). We define D 2 (t) = θ the derivative of θ, which is well-determined! The discussion in Sect , in part. (1.6), shows that K(r) = κ, the curvature function associated to r. Let k : I R denote a continuous function and let s 0 I. Define 1. angle functions θ : I R as θ(s) = s s 0 k(u) du + θ 0 for arbitrary θ 0 R; those are all functions with θ = k. 2. unit vector functions t : I S 1 R 2 by t(s) = [cosθ(s), sin θ(s)] for every θ in 1.; those are the only unit vector functions with k(t) = k. 3. vector functions r : I R 2 by r(s) = [ s s 0 cosθ(u) du +x 0, s s 0 sin θ(u) du + y 0 ] for every θ in 1. and arbitrary [x 0, y 0 ] R 2 ; those are then the only vector functions with K(r) = k. To summarize, any of the parametrizations r is essentially constructed by a double integration of the function k. It follows from the construction that K(r) = k. Uniqueness up to a rigid motion: Let r 1,r 2 denote two parametrizations with K(r 1 ) = K(r 2 ) = k, corresponding to choices θ 0 = θ 1, resp. θ 0 = θ 2, and [x 0, y 0 ] = [x 1, y 1 ], resp. [x 0, y 0 ] = [x 2, y 2 ]. Then, the associated unit vector functions t i are related: In fact, t 2 is obtained from t 1 through a rotation Φ 2 by the angle θ 2 θ 1 around the origin. Note that the rotation Φ 2 is a linear map. Differentiation (and thus integration) commutes with linear maps; this is why the integral of t 2 is still obtainable from the integral t 1 through a rotation by the angle θ 2 θ 1. Finally, letting Φ 1 denote the planar translation given by [x 2 x 1, y 2 y 1 ], the vector function r 2 can be obtained from the vector function r 1 through the rigid motion Φ = Φ 1 Φ 2 of the plane. This result might seem merely academic; but certain curves are constructed in this way for practical purposes. Roads leading to or from a highway frequently consist of parts with curvature increasing (or decreasing) at a constant rate. On such a road, the driver has to turn his/her driving wheel at a constant rate. An example of a function increasing with a constant rate is κ(s) = s. The resulting curve in Fig. 1.6 is called a clothoid curve it has κ(s) = s as the associated curvature function, and you have probably experienced it as a driver!

32 32 CHAPTER 1. CURVES IN PLANE AND SPACE Figure 1.6: Curve with κ(s) = s, 3 s 3 Illustration 1.52 Applet: Parametrization of κ, graph of κ, choice of coordinates of start point and start angle yield curve The local canonical form of a plane curve Locally, a smooth vector function can be approximated by correponding Taylor polynomials. Let us try to rediscover curvature of a plane curve in the 2nd degree Taylor polynomium of a (unit speed) parametrization r : I R 2. For simplicity, assume that 0 int(i), and (possibly after a rigid motion of the plane), that r(0) = 0, that r (0) = t(0) = [1, 0], and that the curve has curvature κ at the origin. As a result of Def. 1.36, we obtain the 2nd derivative r (0) = κ t(0) = [0, κ]. In particular, the 2nd degree Taylor polynomial is of the form r (2) (0)(s) = [0, 0] + [1, 0]s + [0, κ]s2 2 = [s, κ 2 s2 ]. This representation (plus the corresponding error term) is known as the local canonical form of the curve in a neighbourhood of P 0 = O. It represents a parabola with curvature κ at its top point at O. The sign of κ determines, whether this approximating parabola (and the original curve close to P 0 ) is situated in the upper or in the lower half-plane. This corresponds to the result in Cor Illustration 1.53 Curves and 2nd degree Taylor polynomials at various points.

33 1.3. CURVATURE Exercises 1. Given three points P, Q, R R 2, not contained in a line l R 3. Show: (a) There is a unique circle containing P, Q, R. Find the center C(P, Q, R) and the radius ρ(p, Q, R) of this circle. (b) Show that C(P s0, P s1, P s2 ) and ρ(p s0, P s1, P s2 ) converge when s 1 and s 2 tend to s (a) Let C be the (spiral) curve given by the regular parametrization r : ]ε, [ R 2, r(t) = [t cost, t sin t], ε > 0. Determine the velocity r (t), the speed v(t) = r (t) and the unit tangent vector field t(t) associated to this parametrization. Plot the x-coordinate of t(t). Determine an angle function θ :]ε, [ R such that t(t) = [cos(θ(t)), sin(θ(t))]. You cannot get the curvature function immediately by differentiating θ, since r is not a unit speed parametrization. (b) Calculate the associated curvature function κ(t) and show that lim t κ(t) = 0. (c) Determine a parametrization for Cs evolute curve E C. 3. Let C denote a plane curve with unit speed parametrization and associated unit tangent and normal vector fields t(s) and n(s) = ˆt(s). Show (using Prop. 1.10): (a) n (s) n(s) = 0. (b) n (s) t(s) = κ(s). (c) n (s) = κ(s)t(s). 4. (a) Show Prop about evolute curves using Lemma (b) Let C denote a regular plane curve with evolute E C. Assume C given by a unit speed parametrization, and that κ is a smooth function of the parameter s. Show that the derivative of the associated parametrization for E C at C Ps is given by κ (s) n(s). κ(s) (c) Show that E C is regular except at the points C Ps corresponding to points P s C with critical values of κ, i.e., κ (s) = 0. (d) Let P s C and C Ps E C be two associated points on C and its evolute curve E C such that κ (s) 0. Show that the tangent lines to C at P and to E C at C P are perpendicular to each other. (e) An exercise on involutes, e.g., the involute of a circle. 5. Reprove Prop using Thm